HW 12b Help and hint
#2
To fill a child's inflatable wading pool you use a garden hose with a diameter of 2.8 cm. Water
flows from this hose with a speed of 1.1 m/s. How long will it take to fill the pool to a depth of 33
cm if it is circular and has a diameter of 3.8 m?
Someone asks: I'm having trouble relating the velocity of the water exiting the hose to the
volume/area of the pool. Where should I start? Yeah I'm right there with you. I have no idea how
to do this problem!"
My answer:
This question is a perfect example of how to apply what you learned from textbook and classes
to real life questions.
They ask you to find out the time it takes to fill the pool and they told you the size of the pool.
You can immediately realize the pool's volume matters, The total time to fill the pool, would be
the pool's volume divided by the amount of water added in to the pool per second.
Pool volume is straight forward to calculate. But what is the flow rate, Or let's say what is the
amount of water added per second?
The larger the hose penning is, the more the flow rate should be. The fast the water flows, the
more the flow rate should be.
flow rate (amount of water added per second) is = A*v
Remember Why we agreed that A1v1=A2v2 in the class, when I tried to use the window and the
door as the openings for the flow?
The idea is that for the classroom filled with incompressible liquid(water), every second the
amount comes in has to go out.
Every second how much comes in? That would be the opening area of the incoming end A1
multiply v1, the speed of the flow there. That's why A1*v1=A2*v2.
In this question there is no two ends. water only comes into the pool and doesn't leak out, but if
you really understand why A1*v1=A2*v2, you would know that the amount of water you add
though the hose per second is A*v of that hose. See the end of page 493-497.
And the total amount of water you add during time period t is simply
Volume=A*v*t
#3, Part b is not required. But if you are interested, you can try the following:
A water pipe having a 2.6 cm inside diameter carries water into the basement of a house at a
speed of 0.85 m/s and a pressure of 166 kPa. The pipe tapers to 1.2 cm and rises to the second
floor 7.8 m above the input point.
Someone asked:
Can anyone help me figure out part b to this problem? I have tried many different equations, I
keep getting the answer wrong. I've used P2=P1-roeg(y2-y1) and also P2=P1+1/2roe(v1^2-v2^2)
what's wrong with using these equations?
My answers:
Both the two equation you used were not exactly correct.
The 1st equation you used here says the pressure is higher at a deeper location (lower y). (This is
only when fluid is not moving v1=v2=0)
The 2nd equation you used says the pressure is higher where velocity of flow is lower. (This is
when fluid is on very close or same height y1=y2 )
For HW 12 #3, both of the two effects mentioned above coexist. As a result, the correct
equation is
p1+roe* g* y1+ 1/2 roe v1^2 = p2+roe* g* y2+ 1/2 roe v2^2
Someone else then said:
I am also having problems with part b . I used the equation that you gave, but I got it wrong. Rho
can be cancelled, right? Since the density is constant, can't I just use the equation
p1+y1+1/2v1^2=p2+y2+1/2v2^2 ? While designating y1 to be zero? Please help.
My response: Right. You can not cancel the density, look at
p+roe*y+ 1/2 roe v^2
The first item p do not include density directly and only the later two items have density.
When you cancel something in an equation, you need to divide that for EVERY Item in both sides
of your equation.
Someone else then asked
im having trouble finding the velocity in part a. i was able to find the pressure in part b. how did
you go about part a?
In order to know How to find v?
please read my answer to the other question of how to calculate the time of filling the
swimming pool and really understand the equation A1*v1= A2*v2.
(recall that you can narrow the hose end to increase the water speed there when you fight with
your little friend...) How and why?
The speed of flow is directly related to the area of hose opening.
#4 (a) Find the pressure difference on an airplane wing where air flows over the upper
surface with a speed of 121 m/s, and along the bottom surface with a speed of 103 m/s.
(b) If the area of the wing is 37 m2, what is the net upward force exerted on the wing?
How exciting it is to understand the lift force on airplane and be able to calculate it!
In air, density of air is very low: 1.29 kg per m^3
Notice that ρgh is not considered here. Because the h difference is at most a few meters.
And ρ of air is very small (1/1000 of water), so that ρgh across a few meters of height
difference (across the airplane wing thickness) do not change pressure noticeably.
Here we only consider the air pressure difference associated with speed differences. The
pressure is higher where velocity of flow is lower. p1+ 1/2 roe v1^2 = p2+ 1/2 roe v2^2
Notice the unit is kPa or kN. 1kPa =1000Pa